METHODS OF JOINT TRUSS MEMBERS ANALYSIS BASIC INFORMATION
How To Do The Methods Of Joint Truss Members Analysis?
This method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces, and only two independent equilibrium equations are involved.
We begin the analysis with any joint where at least one known load exists and where not more than two unknown forces are present. The solution may be started with the pin at the left end. Its free-body diagram is shown in Fig. 4/7.
With the joints indicated by letters, we usually designate the force in each member by the two letters defining the ends of the member. The proper directions of the forces should be evident by inspection for this simple case.
The free-body diagrams of portions of members AF and AB are also shown to clearly indicate the mechanism of the action and reaction. The member AB actually makes contact on the left side of the pin, although the force AB is drawn from the right side and is shown acting away from the pin.
Thus, if we consistently draw the force arrows on the same side of the pin as the member, then tension (such as AB) will always be indicated by an arrow away from the pin, and compression (such as AF) will always be indicated by an arrow toward the pin.
The magnitude of AF is obtained from the equation ΣFy = 0 and AB is then found from ΣFx = 0. Joint F may be analyzed next, since it now contains only two unknowns, EF and BF. Proceeding to the next joint having no more than two unknowns, we subsequently analyze joints B, C, E, and D in that order.
Figure 4/8 shows the free-body diagram of each joint and its corresponding force polygon, which represents graphically the two equilibrium conditions ΣFx = 0 and ΣFy = 0. The numbers indicate the order in which the joints are analyzed.
We note that, when joint D is finally reached, the computed reaction R2 must be in equilibrium with the forces in members CD and ED, which were determined previously from the two neighboring joints. This requirement provides a check on the correctness of our work.
Note that isolation of joint C shows that the force in CE is zero when the equation ΣFy = 0 is applied. The force in this member would not be zero, of course, if an external vertical load were applied at C.
It is often convenient to indicate the tension T and compression C of the various members directly on the original truss diagram by drawing arrows away from the pins for tension and toward the pins for compression.
This designation is illustrated at the bottom of Fig. 4/8. Sometimes we cannot initially assign the correct direction of one or both of the unknown forces acting on a given pin. If so, we may make an arbitrary assignment. A negative computed force value indicates that the initially assumed direction is incorrect.
How To Do The Methods Of Joint Truss Members Analysis?
This method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces, and only two independent equilibrium equations are involved.
We begin the analysis with any joint where at least one known load exists and where not more than two unknown forces are present. The solution may be started with the pin at the left end. Its free-body diagram is shown in Fig. 4/7.
With the joints indicated by letters, we usually designate the force in each member by the two letters defining the ends of the member. The proper directions of the forces should be evident by inspection for this simple case.
The free-body diagrams of portions of members AF and AB are also shown to clearly indicate the mechanism of the action and reaction. The member AB actually makes contact on the left side of the pin, although the force AB is drawn from the right side and is shown acting away from the pin.
Thus, if we consistently draw the force arrows on the same side of the pin as the member, then tension (such as AB) will always be indicated by an arrow away from the pin, and compression (such as AF) will always be indicated by an arrow toward the pin.
The magnitude of AF is obtained from the equation ΣFy = 0 and AB is then found from ΣFx = 0. Joint F may be analyzed next, since it now contains only two unknowns, EF and BF. Proceeding to the next joint having no more than two unknowns, we subsequently analyze joints B, C, E, and D in that order.
Figure 4/8 shows the free-body diagram of each joint and its corresponding force polygon, which represents graphically the two equilibrium conditions ΣFx = 0 and ΣFy = 0. The numbers indicate the order in which the joints are analyzed.
We note that, when joint D is finally reached, the computed reaction R2 must be in equilibrium with the forces in members CD and ED, which were determined previously from the two neighboring joints. This requirement provides a check on the correctness of our work.
Note that isolation of joint C shows that the force in CE is zero when the equation ΣFy = 0 is applied. The force in this member would not be zero, of course, if an external vertical load were applied at C.
It is often convenient to indicate the tension T and compression C of the various members directly on the original truss diagram by drawing arrows away from the pins for tension and toward the pins for compression.
This designation is illustrated at the bottom of Fig. 4/8. Sometimes we cannot initially assign the correct direction of one or both of the unknown forces acting on a given pin. If so, we may make an arbitrary assignment. A negative computed force value indicates that the initially assumed direction is incorrect.
